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Natalka [10]
3 years ago
8

Kevin claims he can draw a trapezoid with 3 right angles.Is this possible?explain

Mathematics
1 answer:
Pepsi [2]3 years ago
7 0
This is not possible. Since trapezoids are quadrilaterals, the sum of their angles must add up to 360. If three of the angles are right angles (90 degrees each), then the only possible measurement of the last angle would be 90 degrees, making a fourth right angle. With four right angles, this shape would be a square, not a trapezoid.
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What is 3 and 1/3 multiplied by 2 3/5
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The answer to the problem is 8 2/3.

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4 years ago
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(/25) 11111111.11111111.11111111.10000000 dotted decimal subnet mask equivalent: ________________________________ number of subn
hjlf

Count the number of 1 bits after the last decimal point. Call this number n.

There will be 2^n subnets and 2^(8-n) - 2 hosts per subnet. (You can mutilply the number of subnets by the number of hosts per subnet to find the total number of hosts.)

/25: 255.255.255.128, 2 subnets, 126 hosts per subnet

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5 0
3 years ago
What does this graph represent?
grandymaker [24]

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I think its a car coming to a stop.

Step-by-step explanation:

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5 0
4 years ago
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a plane travels at a speed of 180 mph in still air. flying with a tailwind, the plane is clocked over a distance of 700 miles. f
galben [10]
<h3>Answer: approximately 43.57337308309 mph</h3>

Round this value however you need to

===========================================================

Explanation:

w = speed of the wind

The plane travels at a speed of 180 mph in still air. With a tailwind, the plane speeds up (the wind is coming from the tail to push the plane forward more), so the plane is now traveling at a speed of 180+w miles per hour. It travels 700 miles

d = r*t

700 = (180+w)*t

t = 700/(180+w)

This is the time it takes when the wind speeds up the plane.

In contrast, the headwind slows the plane down because now the wind is going against the plane. Headwinds attack the head of the plane and push against the plane. With a headwind, the plane is now going 180-w miles per hour. It travels for 2 hours longer to do the return trip (of 700 miles), so,

d = r*t

700 = (180-w)*(t+2)

700 = 180t + 360 - w*t - 2w

700 = 180( 700/(180+w) ) + 360 - w*( 700/(180+w) ) - 2w

700 = 126000/(180+w) + 360 - 700w/(180+w) - 2w

700(180+w) = 126000 + 360(180+w) - 700w - 2w(180+w)

126000+700w = 126000+64800+360w-700w-360w-2w^2

126000+700w = -2w^2 - 700w + 190800

-2w^2 - 700w + 190800 = 126000+700w

-2w^2 - 700w + 190800 - 126000-700w = 0

-2w^2 - 1400w + 64800 = 0

Use a graphing calculator or the quadratic formula to find the approximate solutions are

w = -743.5733730831, w = 43.57337308309

Ignore the negative value as it makes no sense to have a negative wind speed.

The only practical answer is approximately 43.57337308309 mph

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3 years ago
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